JEE Advance - Chemistry (2019 - Paper 2 Offline - No. 10)
The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm$$-$$3, then molarity of urea solution is ................
(Given data : Molar masses of urea and water are 60 g mol$$-$$1 and 18 g mol$$-$$1, respectively)
(Given data : Molar masses of urea and water are 60 g mol$$-$$1 and 18 g mol$$-$$1, respectively)
Answer
2.98
Explanation
Molarity
$$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$$
Also, volume = $${{Mass} \over {Density}}$$
Given, mole fraction of urea $$({\chi _{urea}})$$ = 0.05
Mass of water = 900 g
Density = 1.2 g/cm3
$${\chi _{urea}}$$ = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$$
[$$ \because $$ Moles of water = $${{900} \over {18}}$$ = 50]
$$0.05 = $$$${{{n_{urea}}} \over {{n_{urea}} + 50}}$$
$$ \Rightarrow $$ $$19{n_{urea}} = 50$$
$${n_{urea}} = 2.6315$$ moles
$${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$$
$$ = (2.6315 \times 60)g$$
$$V = {{2.6315 \times 60 + 900} \over {1.2}}$$
[$$ \because $$ $$Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$$]
= 881.57 mL
Now, molarity
= $$Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$$
$$ = {{2.6315 \times 1000} \over {881.57}}$$ = 2.98 M
$$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$$
Also, volume = $${{Mass} \over {Density}}$$
Given, mole fraction of urea $$({\chi _{urea}})$$ = 0.05
Mass of water = 900 g
Density = 1.2 g/cm3
$${\chi _{urea}}$$ = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$$
[$$ \because $$ Moles of water = $${{900} \over {18}}$$ = 50]
$$0.05 = $$$${{{n_{urea}}} \over {{n_{urea}} + 50}}$$
$$ \Rightarrow $$ $$19{n_{urea}} = 50$$
$${n_{urea}} = 2.6315$$ moles
$${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$$
$$ = (2.6315 \times 60)g$$
$$V = {{2.6315 \times 60 + 900} \over {1.2}}$$
[$$ \because $$ $$Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$$]
= 881.57 mL
Now, molarity
= $$Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$$
$$ = {{2.6315 \times 1000} \over {881.57}}$$ = 2.98 M
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